()22ˆˆ3.0 [(3.0)( 6.0) (8.0)(5.0)]k ( 1.7 10 kg m s)k.=−−=−×⋅GA(b) The torque is given by Eq. 11-14, GGGτ=×rF.We write Grxy=+##ij and GFFx=#i and obtainG=+×=−xyFyFxx##ijikejejsinceii×=0and## #ji k.×=−Thus, we find()8.0m7.0N k (56N m)k.−=⋅G(c) According to Newton’s second law GGA=ddt, so the rate of change of the angular momentum is 56 kg⋅m2/s2, in the positive zdirection. 33. If we write (for the general case) Grxyz=++###k,then (using Eq. 3-30) we find GGrv×is equal to
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.